# trig function?

can somebody just provide me with how to figure this out? i don't need the answers an explanation will be great. Thank you

### 4 Answers

- TomVLv 72 months agoFavorite Answer
Use the relationships:

sinΘ = y/r

cosΘ = x/r

tanΘ = sinΘ/cosΘ = y/x

secΘ = 1/cosΘ = r/x

cscΘ = 1/sinΘ = r/y

cotΘ = 1/tanΘ = x/y

and the fact that in Quadrant 2, y is positive and x is negative.

Given: sinΘ = 2/7 from which we know that r = 7 and y = 2

Calculate x = ±√(r² - y²) [choose the negative value because of Q2]

x = -√(49-4) = -√45 = -3√5

Use those values to calculate the remaining functions.

cosΘ = -3√5/7

tanΘ = 2/(-3√5) = -2√5/15

secΘ = 7/(-3√5) = -7√5/15

cscΘ = 7/2

tanΘ = -3√5/2

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- lenpol7Lv 72 months ago
By Pythagoras we need to find the base(Adjacent side).

Hence

7^2 = 2^2 + b^2

b^2 = 7^2 - 2^2

b^2 = (7 - 2)(7 + 2)

b^2 = 5 * 9 = 45

b = 6.7082...

b = -6.7082 .. is negative because it is in the 2nd quadrant.

Hence

Cos(Th) = 6.7082.../7

Tan(Th) = 2/6.7082...

Sec(th) = 1/Cos(Th) = 7/6.7082...

Csc(Th) = 1/Sin(Th) = 7/2

Cot(Th) = 1/Tan(Th) = 6.7082../2

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- PopeLv 72 months ago
cos²(θ) = 1 - sin²(θ)

cos²(θ) = 1 - (2/7)²

cos²(θ) = 45/49

cos(θ) = -5√(3)/7 ... negative because θ is in quadrant II

tan(θ) = sin(θ)/cos(θ)

sec(θ) = 1/cos(θ)

csc(θ) = 1/sin(θ)

cot(θ) = cos(θ)/sin(θ)

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- AshLv 72 months ago
(a) Remember the identity involving sin and cos....

sin²θ+cos²θ = 1

cos²θ = 1- sin²θ

Now plug in sinθ and find cosθ

b) tanθ = sinθ / cosθ

c) secθ = 1/cosθ

d) cscθ = 1/sinθ

e) cotθ = 1/tanθ

Good luck !

- olayda2 months agoReport
thank you!

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